Answer
$+13.0\times 10^{-6}~C$
Work Step by Step
In part (a), we found that the charge on the cavity wall is $-3.0\times 10^{-6}~C$
The net charge on the conductor is $+10\times 10^{-6}~C$
We can find the charge $Q$ on the outer surface:
$(-3.0\times 10^{-6}~C)+Q = +10\times 10^{-6}~C$
$Q = +13.0\times 10^{-6}~C$
