and now potential energy

$\mathrm{U}=\frac{(\mathrm{k})\left(64 \mathrm{Q}^{2}\right)}{\mathrm{d}^{2}}+\frac{\mathrm{k}\left(16 \mathrm{Q}^{2}\right)}{\mathrm{d}-\mathrm{x}}+\frac{\mathrm{k} 4 \mathrm{Q}^{2}}{\mathrm{x}}$

$\frac{d U}{d x}=0 \Rightarrow \frac{+16}{(d-x)^{2}}=\frac{4}{x^{2}}$

$\Rightarrow \pm 2 x=d-x$

$\Rightarrow x \pm 2 x=d$

$\Rightarrow x=d / 3$ or $x=-d$

$\Rightarrow x=\frac{9}{3}=3 \mathrm{cm}$

Field at $Q$ is

$=\frac{k(4 Q)}{(3 c m)^{2}}-\frac{k(16 Q)}{(6 c m)^{2}}=0$