Know some thermodynamics!

First find the heat capacity ratio/adiabatic factor for the mixture of $He$ and $O_{2}$ . This comes out to be 1.5

Now, since the piston is non conducting , the mixture in the right part undergoes an adiabatic compression .Since its volumes gets halved, the pressure exerted by the mixture can be found using $PV^{1.5} = constant$ . The pressure comes out to be $2\sqrt2P_{0}$ , where $P_{0}$ is initial pressure. Now, the pressure exerted by the gas in the left part can also be found. Note that even the spring exerts a force since its compressed. The pressure of the gas in left part comes out to be $2\sqrt2 P_{0} +~~ \frac{KL}{2A}$ .

Now using the equation $\frac{PV}{T} = constant$ the temperature of the the gas in the left part can be found and thus change in its internal energy can be found. Now, find the work done by the gas in left part. To find this , first find the work done $on$ the gas in the right compartment

Work done by helium in left part = Work done on the mixture in right part + increase in potential energy stored in the spring

Using First law of Thermodynamics ,

Heat supplied to the gas = Change in internal energy + Work done by the gas .

Substituting, we get $\color{#3D99F6}{\boxed {Heat ~ supplied = ~ \frac{(13\sqrt2 -7)P_{0}AL}{2} +~ \frac{10KL^{2}}{8}}}$ .

This comes out to be $149.5$ . Thus, answer is $150$