1. Set up the velocity vectors

Let the river flow along the positive $x$-direction with speed $x$ m/s.

The swimmer’s speed relative to water is $10$ m/s, and he swims at an angle of $120^\circ$ with the flow.

So the swimmer’s velocity relative to water has components:

$v_{sx}=10\cos 120^\circ = 10\left(-\frac12\right)=-5$

$v_{sy}=10\sin 120^\circ = 10\left(\frac{\sqrt3}{2}\right)=5\sqrt3$

Thus,

$\vec v_{s/w}=(-5)\hat i + (5\sqrt3)\hat j$

2. Add river flow velocity

The river velocity is

$\vec v_{r}=x\hat i$

Hence the swimmer’s velocity relative to ground is

$\vec v = \vec v_{s/w} + \vec v_r = (x-5)\hat i + 5\sqrt3\hat j$

3. Condition for reaching directly opposite point

To reach the point directly opposite, there should be no net horizontal drift.

Therefore, the $x$-component of resultant velocity must be zero:

$x-5=0$

So,

$x=5$

4. Nearest integer

The value is already an integer:

$\boxed{5}$

5. Comparison with stored answer

Stored correct answer = $5$

Our derived answer = $5$

So they agree.