Answer
$C = 40~\mu F$
Work Step by Step
We can find an expression for the initial charge on the capacitor:
$q = (100~V)~C$
We can find an expression for the equivalent capacitance when the capacitor is connected across the $60~\mu F$ capacitor:
$C_{eq} = C+60~\mu F$
We can find $C$:
$40~V = \frac{q}{C_{eq}}$
$40~V = \frac{q}{C+60~\mu F}$
$40~V = \frac{(100~V)~C}{C+60~\mu F}$
$(40~V)(C+60~\mu F) =(100~V)~C$
$(60~V)~C = 2400~\mu C$
$C = \frac{2400~\mu C}{60~V}$
$C = 40~\mu F$
