Let $v_1$ be the speed at the lowest point and $v_2$ be the speed at the highest point. By conservation of mechanical energy from the highest to the lowest point: $\frac{1}{2} m v_1^2 = \frac{1}{2} m v_2^2 + mg(2L)$ $v_1^2 = v_2^2 + 4gL$

The maximum tension ($T_{max}$) occurs at the lowest point of the circular path: $T_{max} = \frac{mv_1^2}{L} + mg$ Substitute $v_1^2 = v_2^2 + 4gL$ into the equation: $T_{max} = \frac{m(v_2^2 + 4gL)}{L} + mg = \frac{mv_2^2}{L} + 4mg + mg = \frac{mv_2^2}{L} + 5mg$

The minimum tension ($T_{min}$) occurs at the highest point: $T_{min} = \frac{mv_2^2}{L} - mg$

Given the ratio of maximum to minimum tension is 4: $\frac{T_{max}}{T_{min}} = 4$ $\frac{\frac{mv_2^2}{L} + 5mg}{\frac{mv_2^2}{L} - mg} = 4$ $mv_2^2 + 5mgL = 4(mv_2^2 - mgL)$ $mv_2^2 + 5mgL = 4mv_2^2 - 4mgL$ $3mv_2^2 = 9mgL$ $v_2^2 = 3gL$

Given $g = 10 \text{ m/s}^2$ and $L = \frac{10}{3} \text{ m}$, substitute these values: $v_2 = \sqrt{3 \times 10 \times \frac{10}{3}}$ $v_2 = \sqrt{100} = 10 \text{ m/s}$

Thus, the speed of the stone at the highest point is $10 \text{ m/s}$.