Solution:

Given Data
- Two spherical conductors \( B \) and \( C \) have equal radii and equal charges.
- They repel each other with a force \( F \) when kept at a certain distance apart.

Step-by-Step Solution
1. Initial Charge on B and C:
Let's assume the initial charge on both \( B \) and \( C \) is \( q \).
Since they are at a certain distance \( r \) apart, the initial force of repulsion between \( B \) and \( C \) is given by Coulomb's law:
\[
F = \frac{k q^2}{r^2}
\]

2. Introducing the Third Conductor (A):
- A third conductor \( A \) with the same radius as \( B \) and \( C \) is initially uncharged.
- \( A \) is first brought in contact with \( B \), allowing charges to redistribute.

3. Charge Redistribution (First Contact with B):
- When \( A \) (initially uncharged) is brought into contact with \( B \) (which has charge \( q \)), the charge will distribute equally between \( A \) and \( B \) because they have the same radius.
- After contact, the charge on each (both \( A \) and \( B \)) will be:
\[
\frac{q}{2}
\]

4. Charge Redistribution (Then Contact with C):
- Next, \( A \) (which now has charge \( \frac{q}{2} \)) is brought in contact with \( C \) (which has charge \( q \)).
- The charge will again distribute equally between \( A \) and \( C \) because they have the same radius.
- After contact, the charge on each (both \( A \) and \( C \)) will be:
\[
\frac{q}{2} + \frac{q}{2} = \frac{3q}{4}
\]
- So, now \( C \) has a charge of \( \frac{3q}{4} \), and \( A \) also has \( \frac{3q}{4} \).

5. Final Charges on B and C:
- After removing \( A \), the charges on \( B \) and \( C \) are as follows:
- \( B \) has \( \frac{q}{2} \).
- \( C \) has \( \frac{3q}{4} \).

6. New Force of Repulsion between \( B \) and \( C \):
- The new force \( F' \) between \( B \) and \( C \), separated by the same distance \( r \), is given by:
\[
F' = \frac{k \left(\frac{q}{2}\right) \left(\frac{3q}{4}\right)}{r^2}
\]
- Simplifying, we get:
\[
F' = \frac{k \cdot q^2 \cdot 3}{8r^2} = \frac{3}{8} \cdot \frac{k q^2}{r^2}
\]
- Since \( F = \frac{k q^2}{r^2} \), we can write:
\[
F' = \frac{3}{8} F
\]

Final Answer
The new force of repulsion between \( B \) and \( C \) is:
\[
\frac{3F}{8}
\]