Hi Suraj. In the future, please let us know which class this is for when you make the post.

Let $P_0$ be atmospheric pressure and $P$ denote the pressure at the bottom of the green liquid. Hence we know $$P = P_0 + \rho_g\cdot g\cdot h_2$$ where $\rho_g$ is the density of the green liquid.

This pressure is the same at the other side of the tube at the same height. Hence $$P = P_0 + \rho_b\cdot g\cdot (h_2 - h_1)$$ where $\rho_b = 1000$ is the density of water (the blue liquid).

Setting these equal we can solve for $\rho_g$, this gives $$\rho_g = \rho_b\cdot \frac{h_2 - h_1}{h_2} = 1000\cdot \frac{4.75 - 0.22}{4.75} = 953.68\approx 954$$ in this case.