How can be calculated the open circuit power test on this transformer?
You're on the right track, but we have to do a little more calculation.  The short circuit test data allows us to calculate the resistance.  In the schematic below, we apply some voltage \$V_1\$ and measure the current \$I_1\$ on the primary side.  What we can then calculate is \$Z_1\$, the equivalent series impedance which is \$R_1 + j X_1\$.







simulate this circuit &ndash; Schematic created using CircuitLab

In particular, we have $$R_1 = \frac{P_1}{|I_1|^2}$$  In this case, we have \$R_1=37947/225^2 = 0.75\Omega\$ where \$R_1\$ is the equivalent series resistance on the primary side.  If we want to be pedantic, although it's not needed for this problem, we could calculate \$X_1\$ because we know that $$|Z_1|^2 = \left(\frac{|V_1|}{|I_1|}\right)^2 = R_1^2 + X_1^2$$
To translate that to a secondary side equivalent series resistance, we multiply by the square of the turns ratio: \$R_2=R_1\left(\frac{50000V}{10000V}\right)^2=18.739\Omega\$. 







simulate this circuit

There are a number of equivalent ways to express efficiency, but in this case the most useful form is $$\eta=\frac{P_{out}}{P_{out}+W_C+W_I}$$ where \$W_C\$ is the copper loss and \$W_I\$ are hysteresis and eddy losses, which are usually combined and called "iron losses" or "core losses".  The other key thing to remember is that iron losses are not dependent on load current while copper losses are: \$W_C = |I|^2R\$  Substituting that into the equation gives $$\eta=\frac{P_{out}}{P_{out}+|I_2|^2R_2+W_I}$$  A little algebra lets us solve for \$W_I\$. $$W_I=\frac{P_{out}}{\eta}-P_{out}-|I_2|^2R_2$$
Since we're given efficiency at unity power factor at nominal we can simply plug in the numbers and calculate the result:
$$W_I = \frac{2500 kW}{0.9766} - 2500kW - (50A)^2(18.739\Omega)$$
Here, the 50A is the secondary current at nominal (2500VA/50kV).  We could equivalently used the primary current at nominal (250A = 2500VA/10kV) and \$R_1\$.  Mathematically, they're identical.
I calculate \$W_I = 13.054kW\$ which is, directly, the open circuit power.  

One can also apply this the other way around.  That is, if we are given the open circuit power, we have \$W_I\$ and can, given the efficiency \$\eta\$, calculate \$W_C\$.  Since we know that this is at the rated current, which is also how the short circuit test is conducted (if you can forgive the pun) we can extract \$R2\$.