Angular diameter of the Sun's reflection from the ocean, seen from Sun-Earth L1?
$d_o$ and $h_o$ are the distance and diameter of the Sun and $d_i$ and $h_i$  of the respective images. The Earth's radius is $r$ and $f=\frac{r}{2}$ is the focal length.

The mirror equation $$\frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f}$$ in the approximation $d_o \gg f$ gives $d_i \approx f$. Then, using the magnification equation

$$h_i = \frac {d_i}{d_o}h_o \approx \frac {f}{d_o}h_o.$$ 

If $S$ is the distance to the satellite then the angular size is

$$\frac{h_i}{S} \approx \frac{h_o}{d_o}\frac{f}{S} = \frac{h_o}{d_o}\frac{r}{2S} \approx \frac{1.4\times10^6}{150\times10^6}\frac{6.4\times10^3}{2\times 1.6\times 10^6}\approx 1.8 \times 10^{-5}$$ 

where distances are in km. (We have also ignored that the image is $\frac{R}{2}$ closer to the satellite than is the Earth's center.)

On a more descriptive note, the reflection on perfectly smooth water would look like a very bright spot about 29 km in diameter. It is blurred out (due to sea waves) to about ten times that diameter on the image. It would be interesting to calculate the brightness (per area) of the (unblurred) spot compared to the brightness of the sunlit Earth's surface.