then let $v_{\max }=$ the $\max$

voltage across the capacitors

So charge $=\mathrm{Q}=\mathrm{Cv}_{\max }=\frac{16}{3} \times \mathrm{v}_{\max }$

Now ${v_{1}}=\frac{16}{3} \times v_{\max } \times \frac{1}{16}\left[\left(\frac{Q}{C}\right)\right]$

$=\frac{v_{\max }}{3}<80 \Rightarrow v_{\max }<240 \mathrm{V}$

and $v_{2}=\frac{16}{3} \times v_{\max } \times \frac{1}{8}$

$=\frac{2}{3} v_{\max }<20 \Rightarrow v_{\max }<30 v$

so $v_{\max }=30 \mathrm{V}$

Thus $Q=\frac{16 \times 30}{3}=160 \mu C$