Let the smaller force be $F$. Then the larger force is $3F$.

If the angle between the two forces is $\theta$, then the magnitude of the resultant is $R=\sqrt{F^2+(3F)^2+2(F)(3F)\cos\theta}$ $R=\sqrt{10F^2+6F^2\cos\theta}$

Given that the resultant is equal to the larger force, so $R=3F$

Squaring both sides: $10F^2+6F^2\cos\theta=9F^2$

Divide by $F^2$: $10+6\cos\theta=9$ $6\cos\theta=-1$ $\cos\theta=-\frac{1}{6}$

The angle is given as $\theta=\cos^{-1}\left(\frac{1}{n}\right)$ so $\frac{1}{n}=-\frac{1}{6}$ $n=-6$

Therefore, $|n|=6$