Answer
$a=2.03\times 10^{-5}\frac{m}{s^2}$
Work Step by Step
We know that:
$F=qvB$
We also know that:
$B=\frac{\mu_{\circ}I}{2\pi r}$
$\implies F=qv\frac{\mu_{\circ}I}{2\pi r}$
As $a=\frac{F}{m}$;
$\implies a=\frac{qv\frac{\mu_{\circ}I}{2\pi r}}{m}=\frac{qv}{m}\frac{\mu_{\circ}I}{2\pi r}$
We plug in the known values to obtain:
$a=\frac{18.0\times 10^{-3}\times 3.4\times 4\pi \times 10^{-7}\times 25}{2\pi \times 0.175\times 0.086}=2.03\times 10^{-5}\frac{m}{s^2}$