The shortest wavelength in the continuous X-ray (bremsstrahlung) spectrum is set by the Duane-Hunt law, where the entire electron energy $eV$ appears as one photon: $\lambda_{\min} = hc/(eV)$.
Work in SI units. With $h = 6.63 \times 10^{-34}$ J s, $c = 3 \times 10^8$ m/s, $e = 1.6 \times 10^{-19}$ C and $V = 3 \times 10^4$ V:
\[ \lambda_{\min} = \frac{(6.63 \times 10^{-34})(3 \times 10^8)}{(1.6 \times 10^{-19})(3 \times 10^4)}. \]
The numerator is $1.989 \times 10^{-25}$ and the denominator is $4.8 \times 10^{-15}$, so
\[ \lambda_{\min} = 4.14 \times 10^{-11}\ \text{m} = 41.4\ \text{pm}. \]
Note that $1$ pm $= 10^{-12}$ m, so this is deep in the X-ray band, which rules out the nm, angstrom and micrometre choices.
\[\boxed{\lambda_{\min} \approx 41.4\ \text{pm}}\]