$K=5$
In KNN, we assign the test data point to that class which has the majority representative.
Since, there are $3$ squares and $2$ circles, So, assuming the Euclidean distance or $L_2$-norm distance measure, if we have to assign the test data point i.e. the diamond shaped data point to the square shaped data point, we have to take all the square shaped data points in its neighbor boundary which automatically includes the circle shaped data points.
If we make this setting then we will be having $5$ nearest neighbor data points and so $K=5$ and in this way test data point is assigned the label as sqaure because $\mathbb{P}(\frac{3}{5})>\mathbb{P}(\frac{2}{5})$