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Given:
Masses, $m_1 = m_2 = m_3 = m_4 = m = 1 \text{ kg}$
Radius of the circle, $R = 1 \text{ m}$
Gravitational constant, $G = 6.673 \times 10^{-11} \text{ N m}^2\text{kg}^{-2}$

Gravitational potential is a scalar quantity. The total gravitational potential $V$ at the center $O$ of the circle due to the four masses is:
$V = V_1 + V_2 + V_3 + V_4$
$V = -\frac{Gm_1}{R} - \frac{Gm_2}{R} - \frac{Gm_3}{R} - \frac{Gm_4}{R}$
$V = -4\frac{Gm}{R}$
$V = -4 \times \frac{6.673 \times 10^{-11} \times 1}{1} \text{ J/kg}$
$V = -2.67 \times 10^{-10} \text{ J/kg} \quad (\text{or } -4G \text{ J/kg})$

Answer: $-2.67 \times 10^{-10} \text{ J/kg}$ (or $-4G \text{ J/kg}$)