The value of R is ___ meter.

The potential at any point $$P(x,y)$$ due to two point charges is the algebraic sum of the individual potentials:

$$V(x,y)=k\left(\frac{-Q}{r_1}+\frac{Q/\sqrt{3}}{r_2}\right)$$
where

$$r_1=\sqrt{x^{2}+y^{2}}$$ is the distance of $$P$$ from the charge $$-Q$$ placed at the origin, and

$$r_2=\sqrt{(x-2)^{2}+y^{2}}$$ is the distance of $$P$$ from the charge $$\dfrac{Q}{\sqrt3}$$ placed at the point $$(2,0)$$.

For the given equipotential circle, the potential is zero, so

$$k\left(\frac{-Q}{r_1}+\frac{Q/\sqrt{3}}{r_2}\right)=0\quad\Rightarrow\quad \frac{1}{\sqrt3\,r_2}=\frac{1}{r_1}\quad\Rightarrow\quad r_1=\sqrt3\,r_2 \; -(1)$$

Substitute the expressions for $$r_1$$ and $$r_2$$ from above:

$$\sqrt{x^{2}+y^{2}}=\sqrt3\;\sqrt{(x-2)^{2}+y^{2}}$$

Square both sides to remove the square roots:

$$x^{2}+y^{2}=3\bigl[(x-2)^{2}+y^{2}\bigr]$$

Expand and simplify:

$$x^{2}+y^{2}=3\bigl(x^{2}-4x+4+y^{2}\bigr)$$
$$x^{2}+y^{2}=3x^{2}-12x+12+3y^{2}$$
$$0=2x^{2}-12x+12+2y^{2}$$
Divide by $$2$$:

$$x^{2}-6x+6+y^{2}=0$$

Complete the square for the $$x$$-term:

$$(x^{2}-6x+9)+y^{2}=9-6$$
$$(x-3)^{2}+y^{2}=3$$

This is the standard form of a circle:

Centre: $$(b,0)=(3,0)$$
Radius: $$R=\sqrt3\approx1.732\text{ m}$$

Thus, the required radius lies in the interval 1.70 m to 1.80 m.

Answer: $$R\approx1.73\text{ m}$$ (within 1.70 - 1.80).