Answer
${\bf 7.45 \times 10^{-12}}\;\rm J$
Work Step by Step
We know that
$$
E_{2\gamma} = \frac{1}{2} E_{1\gamma}
$$
Since In the two-photon absorption, the photon energy $ E_{2\gamma} $ will be half the energy of the photon in one-photon absorption.
Assuming that the one-photon absorption photon energy is $ E_\gamma $,
$$
E_{2\gamma} = \frac{1}{2} E_{\gamma}\tag 1
$$
For the one-photon absorption photon, the energy is given by
$$
E_\gamma = \frac{hc}{\lambda}
$$
Plug into (1);
$$
E_{2\gamma} = \frac{1}{2}\frac{hc}{\lambda}
$$
Plug the known;
$$
E_{2\gamma} = \frac{(6.63\times 10^{-34})(3\times 10^8)}{2(420\times 10^{-9})} = {\bf 1.48} \, \text{eV}=\bf2.37 \times 10^{-19} \;\rm J
$$
We know that the energy in each pulse $ E_{\text{pulse}} $ is given by
$$
E_{\text{pulse}} \geq \pi r^2 I \tau E_{2\gamma}
$$
where $ I= 10^{32} \, \text{photons} / \text{m}^{2} \cdot \text{s}^{-1} $ is the minimum intensity needed for two-photon absorption, $r=1.0\,\rm \mu m$ is the beam radius, $\tau=100\;\rm fs$ is the lifetime of the pulse.
So the minimum energy is then
$$
E_{\text{pulse}}= \pi r^2 I\tau E_{2\gamma}
$$
Now, substitute all known;
$$
E_{\text{pulse}}= \pi ( 1.0 \times 10^{-6})^2 \left( 1.0 \times 10^{32} \right) ( 100 \times 10^{-15} ) (2.37 \times 10^{-19} )
$$
$$
E_{\text{pulse}} = \color{red}{\bf 7.45 \times 10^{-12}}\;\rm J
$$
