b
We know that work done is
$W =\int PdV\quad \ldots (1)$
$\Rightarrow P=\frac{ nRT }{ V }\quad \ldots (2)$
$\Rightarrow W =\int \frac{ nRT }{ V } dv\quad \ldots (3)$
and $V = KT ^{2 . / 3}\quad \ldots (4)$
$\Rightarrow W =\int \frac{ nRT }{ KT ^{2 / 3}} \cdot dv \quad \ldots (5)$
$\Rightarrow$ from (4)$: dv =\frac{2}{3} KT ^{-1 / 3} dT$
$\Rightarrow W =\int_{ T _{1}}^{ T _{2}} \frac{ nRT }{ KT ^{2 / 3}} \frac{2}{3} K \frac{1}{ T ^{1 / 3}} dT$
$\Rightarrow W =\frac{2}{3} n R \times\left( T _{2}- T _{1}\right) \ldots(6)$
$\Rightarrow T _{2}- T _{1}=90 K \ldots(7)$
$\Rightarrow W=\frac{2}{3} n R \times 90$
$\Rightarrow W =60 nR$
Assuming $1$ mole of gas
$n =1$
So $W =60 \,R$