The heat absorbed by a system in going through the given cyclic process is :

For a cyclic process,

ΔU=0

So from first law,

Q=W

Thus heat absorbed equals net work done, which is area enclosed by the cycle on the P-V diagram.

The loop is a circle touching

P=340,  P=60 and

V=340,  V=60

So diameter is

$$340−60=280$$

Hence radius is

r=140

Area enclosed

$$W=\pi r^2$$

$$=\pi(140)^2$$

$$=19600\pi$$

Now units:

Pressure is in cc and volume in kPa (as labeled), so

$$1(\text{kPa})(\text{cc})=10^3\times10^{-6}=10^{-3}J$$

Thus

$$Q=19600\pi\times10^{-3}$$

$$=19.6\pi\ \text{J}=61.6J$$