Step 1: Use parity to narrow down the states.
Each oscillator eigenstate $|n\rangle$ has a definite parity under $x \to -x$, equal to $(-1)^n$. The ground state $|0\rangle$ is even. The perturbation $V = \lambda x^3$ is an odd function of $x$, since $(-x)^3 = -x^3$.
For the matrix element $\langle m|V|0\rangle$ to survive, the full integrand must be even overall, because an odd integrand over all space integrates to zero. Since $|0\rangle$ is even and $V$ is odd, $|m\rangle$ must be odd, so $m$ must be an odd integer: $m = 1, 3, 5, 7, \dots$ This already rules out option (C), $m=2$, since 2 is even.
Step 2: Count how many quanta $x^3$ can actually change.
Write $x$ in terms of raising and lowering operators, $x \propto (a + a^{\dagger})$. Then $x^3 \propto (a+a^{\dagger})^3$, which expands into 8 terms, each a product of exactly 3 operators, and each operator is either $a$ (changes the quantum number by $-1$) or $a^{\dagger}$ (changes it by $+1$).
Adding three numbers, each equal to $+1$ or $-1$, can only give a total of $+3$, $+1$, $-1$, or $-3$; there is no way to get a net change of $\pm 5$. So acting on $|0\rangle$, the operator $x^3$ can only reach states with $m - 0 \in \{+3, +1, -1, -3\}$. Since $m \geq 0$, only $m = 1$ and $m = 3$ are physically allowed, ruling out option (D), $m=5$.
Step 3: Confirm both surviving states actually appear.
Combining the two results, $m$ must be odd (Step 1) and reachable within 3 steps of quantum number 0 (Step 2). The only values that satisfy both conditions are $m=1$ and $m=3$. A direct expansion of $(a+a^{\dagger})^3|0\rangle$ confirms both terms appear with non-zero coefficients, so neither is accidentally zero.
Final Answer:
Only $m=1$ and $m=3$ give a non-zero matrix element.
\[ \boxed{|m=3\rangle \text{ and } |m=1\rangle} \]