Each of three blocks P, Q and R shown in figure has a mass of 3kg ....

Acceleration of the system, $a = \dfrac{3g}{9} = \dfrac{10}{3}m{s}^{- 2}$

Now, for $R$ we can write

$30 - {T}_{1} = 3 \times a$

$⇒ {T}_{1} = 20N$

Therefore, strain developed in wire $B$, 

 $\text{strain} = \dfrac{\text{stress}}{Y} = \dfrac{\dfrac{{T}_{1}}{A}}{Y} = \dfrac{\dfrac{20}{0 . 005 \times {10}^{- 4}}}{2 \times {10}^{11}}$

$= 2 \times {10}^{- 4}$