Answer
$57.7^{\circ}$
Work Step by Step
The magnetic fore acting on a particle having charge $q$ and moving with a velocity $(\vec v)$ in a uniform magnetic field $(\vec B)$ is given by
$\vec F=q(\vec v\times \vec B)$
or, $|\vec F|=qvB\sin\theta$, where $\theta$ is angle between $\vec v$ and $\vec B$.
In first case, $|\vec F|=F$ and $\theta=25^{\circ}$
Therefore $ F=qvB\sin25^{\circ}$ ..................$(1)$
Let, the particle is moving with an angle $\phi$ with respect to this field, when it experiences the magnetic force.
Thus, in second case,
$2 F=qvB\sin \phi$ ..................$(2)$
Dividing euation $(2)$ by equation $(1)$, we get,
$\frac{2F}{F}=\frac{qvB\sin \phi}{qvB\sin25^{\circ}}$
Or, $ \sin \phi=2\times\sin25^{\circ}$
$\phi\approx57.7^{\circ}$
Thus, the particle will make $57.7^{\circ}$ with the magnetic field, when it experiences a magnetic force of $2F$.
