One mole of a monatomic ideal gas undergoes the process  $A\to B$  in the given $P-V$ diagram. Specific heat capacity in the process is 

From fig. using  $pV=nRT$
At  $A,T_A=\frac{3P_0{V}_0}{R}$
At  $B,T_B=\frac{6P_{0}5V_0}{R}=\frac{30P_0{V}_0}{R}$
 $\Delta U=n{C}_v\Delta T$
For mono-atomic
 $\Delta U=\frac{3R}{2}\Delta T$
We know work done under  $PV$ graph is from  $A\to B$
 $W=3P_0\left(V_0\right)+\frac{1}{2}\left(4V_0\right)\left(3P_0\right)$
 $W=18P_0{V}_0=18\frac{\left[R\Delta T\right]}{27}$
 $W=\frac{2}{3}R\Delta T$

From I law 
 
 $\Delta Q=\Delta U+\Delta W$
 $\Delta Q=3/2R\Delta T+2/3R\Delta T$
 
 $\Delta Q=\frac{13}{6}R\Delta T$
 $\frac{\Delta Q}{\Delta T}=\frac{13}{6}R$

$\therefore C=\frac{1}{n}\frac{\Delta Q}{\Delta T}$
$n=1mole$

$\therefore C=\frac{13}{6}R$