The problem states that the LCM of 12, 48, and m is 336, and their GCF is 4. We need to find the possible value(s) of m.

Let the prime factorization of $m$ be $2^x \times 3^y \times 7^z$.

For the GCF, the exponent of each prime factor is the minimum of the exponents in 12, 48, and $m$.
For prime factor 2: $\min(2, 4, x) = 2$. This implies $x \ge 2$.
For prime factor 3: $\min(1, 1, y) = 0$ (since $3^0$ is in the GCF). This implies $y = 0$.
For prime factor 7: $\min(0, 0, z) = 0$ (since $7^0$ is in the GCF). This implies $z \ge 0$.

For the LCM, the exponent of each prime factor is the maximum of the exponents in 12, 48, and $m$.
For prime factor 2: $\max(2, 4, x) = 4$. This implies $x \le 4$.
For prime factor 3: $\max(1, 1, y) = 1$. This implies $y \le 1$.
For prime factor 7: $\max(0, 0, z) = 1$. This implies $z = 1$.