Given,

The magnitude of a force, $\left|\overrightarrow{F}\right|=\frac{mg}{9}$ and

Force vector makes a varying angle $\theta$ with the horizontal as $\theta =Cx$,

where, $C=10\left(\frac{ \mathrm{degree} }{\mathrm{meter} }\right)$.

Thus, Horizontal component of force, 

$F_x=F \cos \left(\theta \right)=F \cos \left(Cx\right)$.

Applying work energy theorem,

$W_{\mathrm{net}}=∆K.E.$

$\Rightarrow {\int }_0^x{F}_{x}dx=\frac{1}{2}m{v}^2$

(when $\theta$ becomes $30°$, $x=\frac{\theta }{C}=\frac{30°}{10° {\mathrm{m}}^{-1}}=3 \mathrm{m}$)

$\Rightarrow {\int }_0^x\frac{mg}{9} \cos \left(Cx\right) dx=\frac{1}{2}m{v}^2$

$\Rightarrow \frac{mg}{9}{\int }_0^3\cos \left(Cx\right)dx=\frac{1}{2}m{v}^2$

$\Rightarrow \frac{mg}{9C} \left[\sin \left(C \left(3\right)\right)-\sin \left(0\right)\right]=\frac{1}{2}m{v}^2$

$\Rightarrow \frac{10}{9 \left(10° {\mathrm{m}}^{-1}\right)} \left[\sin \left(30°\right)\right]=\frac{1}{2}{v}^2$

$\Rightarrow \frac{1}{9}\left[\frac{1}{2}\right]=\frac{1}{2}{v}^2$

$\Rightarrow v=\frac{1}{3}=0.33 \mathrm{m} {\mathrm{s}}^{-1}$.