The distance between Sun and Earth is $1.6 \times 10^{11} \mathrm{~m}$ and the radius of Earth is $6.4 \times 10^6 \mathrm{~m}$. The ratio of the angular momentum of Earth around the Sun to the angular momentum around its own axis is approximately (Assume Earth as a solid sphere with uniform mass density and rotates around the Sun in a circular path.)

Given, the distance between the Sun and Earth, $r_1=1.6 \times 10^{11} \mathrm{~m}$ Radius of earth, $R_e=6.4 \times 10^6 \mathrm{~m}$ Angular momentum of Earth around the Sun, $L_1=M_e v_1 r_1=M_e \cdot \frac{2 \pi r_1}{T_1} \cdot r_1$ $L_1=2 \times 3.14 \times 6.0 \times 10^{24} \times \frac{\left(1.6 \times 10^{11}\right)^2}{365 \times 24 \times 60 \times 60}$ $\left(\because\right.$ Mass of earth, $M_e=6.0 \times 10^{24} \mathrm{~kg}$ and $T_1=365$ days) $L_1=3.06 \times 10^{40} \mathrm{~kg}-\mathrm{m}^2 / \mathrm{s}$ Angular momentum of Earth about its own axis, where, moment $L_2=I \omega$ of inertia of Earth (uniform solid sphere) around its own axis, $I=\frac{2}{5} M_e R_e^2$. or $\begin{aligned} I & =\frac{2}{5} \times 6.0 \times 10^{24} \times\left(6.4 \times 10^6\right)^2 \\ & \simeq 9.8 \times 10^{37} \mathrm{~kg}-\mathrm{m}^2\end{aligned}$ The earth's angular velocity, $\omega=\frac{2 \pi}{1 \text { day }}$ $=\frac{2 \pi}{24 \times 60 \times 10}=7.3 \times 10^{-5} \mathrm{~s}^{-1}$ $\therefore \quad L_2=9.8 \times 10^{37} \times 7.3 \times 10^{-5} \approx 7.2 \times 10^{33} \mathrm{~kg}-\mathrm{m}^2 \mathrm{~s}^{-1}$ From Eqs. (i) and (ii), we get $\frac{L_1}{L_2}=\frac{3.06 \times 10^{40}}{7.2 \times 10^{33}}=4.3 \times 10^6$ The ratio of angular momentum of Earth around the Sun to the angular momentum around its own axis is $4.3 \times 10^6$ approximately. Hence, the correct option is (c).