A rectangular field is 16 m long and 10 m wide.

Let the width of the path be $x \mathrm{~m}$.

$\therefore$ Length of the field including the path $=16+x+x=16+2 x$

Breadth of the field including the path $=10+x+x=10+2 x$

Now,

(Area of the field including path) $-$ (Area of the field excluding path) $=$ Area of the path

$\Rightarrow(16+2 x)(10+2 x)-(16 \times 10)=120$

$\Rightarrow 160+32 x+20 x+4 x^{2}-160=120$

$\Rightarrow 4 x^{2}+52 x-120=0$

$\Rightarrow x^{2}+13 x-30=0$

$\Rightarrow x^{2}+(15-2) x+30=0$

$\Rightarrow x^{2}+15 x-2 x+30=0$

$\Rightarrow x(x+15)-2(x+15)=0$

$\Rightarrow(x-2)(x+15)=0$

$\Rightarrow x-2=0$ or $x+15=0$

$\Rightarrow x=2$ or $x=-15$

$\Rightarrow x=2 \quad(\because$ Width cannot be negative)

Thus, the width of the path is $2 \mathrm{~m}$.