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 \(W =60 \,R\)