The dimensional formula for strain is same as that of
A. Thrust
B. Angle
C. Modulus of elasticity
D. Stress

Hint: All physical quantities can be derived from the base quantities. In order to find a physical quantity that has some dimensions as that of strain, use elimination method by converting each given option into the fundamental quantities. Afterwards compare it with the dimensional formula for strain to find the correct answer.
Formula Used:
Thrust = Force,$F = {\text{mass}} \times {\text{acceleration}}$
Modulus of elasticity ${\text{ = }}\dfrac{{{\text{Stress}}}}{{{\text{Strain}}}}$
Stress ${\text{ = }}\dfrac{{{\text{Force}}}}{{{\text{Area}}}}$
Strain$ = \dfrac{{\Delta L}}{L}{\text{ }}\left[ {{\text{for linear strain}}} \right]$

Complete step by step answer:
Thrust is defined as the force which is responsible for the movement of an object. In other words thrust is a form of force only. Hence it will have the same dimensions as that of force. Force is defined as the product of mass and acceleration of a body.
Mathematically,
$\eqalign{
  & F{\text{ }} = {\text{ mass}} \times {\text{acceleration}} \cr
  & F = M \times \dfrac{L}{{{T^2}}} \cr
  & \Rightarrow F = \left[ {{M^1}{L^1}{T^{ - 2}}} \right] \cr} $
Hence the above equation gives the dimensional formula of thrust.

The angle is considered to be a dimensionless physical quantity as it is measured in radians and thus is the ratio of two lengths.

Modulus of elasticity is defined as the measure of the ability of an object or a body to resist deformation. Mathematically, it is defined as the ratio of stress and strain.
${\text{Y = }}\dfrac{{{\text{Stress}}}}{{{\text{Strain}}}}{\text{ }}\left[ {{\text{linear}}} \right]$
Now, strain is defined as Force acting per unit area of a body and strain as the ratio of change in dimensions of a body under the action of a force and original dimensions.
So, we have:
$\eqalign{
  & Y = \dfrac{{\dfrac{F}{A}}}{{\dfrac{{\Delta L}}{L}}}{\text{ = }}\dfrac{{FL}}{{A\Delta L}}{\text{ }} \cr
  & {\text{Y = }}\dfrac{{{{\text{M}}^1}L{T^{ - 2}}{L^1}}}{{{L^2}{L^1}}} = {M^1}{L^{ - 1}}{T^{ - 2}} \cr
  & Y = \left[ {{M^1}{L^{ - 1}}{T^{ - 2}}} \right] \cr} $
From the above calculation we can also find that, stress is given in its dimensional formula as:
$\eqalign{
  & \dfrac{{{\text{Force}}}}{{{\text{Area}}}} = \dfrac{{ML{T^{ - 2}}}}{{{L^2}}} \cr
  & \Rightarrow \left[ {M{L^{ - 1}}{T^{ - 2}}} \right] \cr} $
Additionally, it is clear that strain is a dimensionless quantity because it is a ratio of two similar quantities, namely lengths, area or volume.
Therefore, the correct option is B. i.e., angle has the same dimensional formula as strain because both of them are dimensionless.

Note: Although dimensional formula is a very useful way to deduce certain relations between two physical quantities but still has its limitations. For example numeric constants have no dimensions so cannot be reduced using this method.