Recived the same homework and solved as follows:
P=kA ∆T/L
Q/t=kA ∆T/L
Q_1=t.kA ∆T/L=3600 .0,004 .2A=28,8cal/〖cm〗^2
Q_2=L .m=333kJ/kg=79,55cal/g
p=m/v=0,92g/〖cm〗^3
∴Q_2=73,19cal/〖cm〗^3 (substitute density intoheat of transformation)
but Q_1=Q_2 so then cm=28,8/79,55=0,4cm