Standard electrode potential are:
$F{e^{ + 2}}$/$Fe$ $E = $$0.44$
$F{e^{ + 3}}$/$F{e^{ + 2}}$ $E = $$0.77$
If $F{e^{ + 2}}$,$F{e^{ + 3}}$ and $Fe$ block re kept together, then:
A. $F{e^{ + 3}}$increases
B. $F{e^{ + 3}}$ decreases
C. $\dfrac{{F{e^{ + 2}}}}{{F{e^{ + 3}}}}$ remains unchanged
D. $F{e^{ + 2}}$ decreases

Hint: As we know that when a metallic rod is dipped in its own solution, a charge is developed after some time over it with respect to the solution and a potential difference is generated which is known as the electrode potential.

Complete step by step solution:
Let us revise the electrode potential which is an electromotive force generated when a metallic rod is dipped in its own solution, a charge is developed after some time over it with respect to the solution. When the temperature is fixed to ${25^ \circ }C$, pressure is fixed at $1\,atm$ and concentration of solution is also fixed at $1\,M$, in such situations the value of electrode potential is called the Standard electrode potential.
Also we should know that when electrode potential of all elements are arranged in order of their decreasing oxidation potential and increasing order of reduction potential, the series so formed is called an Electrochemical series. There are some properties of this electrochemical series which we should keep in mind and these characteristics are:
Among the known elements Lithium has the highest value of reducing power and thus behaves as the strongest reducing agent due to its high oxidation potential.
Fluorine is having the highest value of oxidizing power and behaves as a strong oxidizing agent due to high reduction potential.
Metal reactivity is directly proportional to the oxidation potential, moving down the group value of oxidation potential decreases hence reactivity decreases. Metals that are more reactive can displace the metal which is less reactive from their salt solutions.
Reactivity of non-metals is directly proportional to reduction potential, moving down the group reduction potential increases hence reactivity of non-metals increases and more reactive non-metal displaces less reactive non-metal from their salt solution.
Therefore we can say that the metals possessing higher negative values of their electrode potential can displace the metals having lower value from their salt solution hence $F{e^{ + 3}}$ will be converted spontaneously to $F{e^{ + 2}}$ and will decrease.

Hence, the correct answer is option (B).

Note:
Remember that stability of metallic oxide and metal nitrate is directly proportional to oxidation potential and moving down the group value of oxidation potential decreases hence stability decreases.