A

Step 1: Calculate the potential due to the inner sphere at point P.
Using the formula for the potential due to a uniformly charged sphere, the potential at a distance \( r > R \) from a charged sphere is given by:
\[ V = \frac{kQ}{r} \]
where \( k \) is Coulomb's constant, \( Q \) is the charge, and \( r \) is the distance from the charge.
For the inner sphere of charge \( 1 \mu C \), the potential at point P (3R) is:
\[ V_1 = \frac{k \times 1 \times 10^{-6}}{3R} \]

Step 2: Calculate the potential due to the outer sphere at point P.
The outer sphere has a charge of \( 2 \mu C \), thus:
\[ V_2 = \frac{k \times 2 \times 10^{-6}}{3R} \]

Step 3: The total potential at point P is the sum of both potentials:
\[ V_{total} = V_1 + V_2 = \frac{k \times 1 \times 10^{-6}}{3R} + \frac{k \times 2 \times 10^{-6}}{3R} \] \[ = \frac{k \times 3 \times 10^{-6}}{3R} = \frac{k \times 10^{-6}}{R} \]

Step 4: We know from the given information that the total potential at point P is 9000 V.
Thus, equating the formulas, we have:
\[ \frac{k \times 10^{-6}}{R} = 9000 \]

Step 5: Solving for R:
\[ R = \frac{k \times 10^{-6}}{9000} \]
Taking \( k = 9 \times 10^9 \, N m^2/C^2 \):
\[ R = \frac{9 \times 10^{9} \times 10^{-6}}{9000} = 1 \, m \]
Therefore, the value of R is 1 meter.