a b c d

d

As the capacitors are in series so the charge on both remains same. Charge on small capacitance $4\mu F$  is Q=4×10−6×100=400μC. Thus we can not apply charge greater than $400\mu C$. The potential of capacitance $10\mu F$ for this charge is $V=400/10=40V$. Thus the maximum potential applied is $(100+40)=140V$