$\therefore$ Total Energy $=\frac{1}{2} KA ^{2}$

By energy conservation we can final $v$ at $x =5$

$\frac{1}{2} K (10)^{2}=\frac{1}{2} K (5)^{2}+\frac{1}{2} mv ^{2}$

$V =\sqrt{\frac{75 K }{ m }}$

Now, velocity is tripled through external mean so the amplitude of $SHM$ will charge and so the total energy, (but potential) energy at this moment will remain same)

$\therefore \frac{1}{2} K (5)^{2}+\frac{1}{2} m \left(3 \sqrt{\frac{75 K }{ m }}\right)^{2}=\frac{1}{2} KA ^{2}$

$\Rightarrow 25 K +675 K = KA ^{2}$

$\therefore A =\sqrt{700}$

$\therefore x =700$