The given square is divided into $$6$$ equal strips in both directions, so the grid contains $$7$$ vertical and $$7$$ horizontal lines.

Hence the number of intersection points is $$7 \times 7 = 49$$. These are the points $$A_1, A_2, \ldots , A_{49}$$.

Two points are called friends when they are directly adjacent in the same row or in the same column, that is, when they share a unit-length edge of the grid.

For a randomly chosen point let the random variable $$X$$ count its number of friends. We have to find $$7E(X)$$.

Instead of evaluating each probability $$p_i$$ separately, we count the total number of “point-friend” incidences in the whole grid and then divide by the number of points.

An unordered adjacent pair contributes $$1$$ friend to each of the two points that form it; thus every such pair contributes $$2$$ to the total $$\sum_{k=1}^{49} X(A_k).$$

Counting adjacent pairs
Horizontal pairs: every one of the $$7$$ rows has $$6$$ adjacent pairs, giving $$7 \times 6 = 42.$$
Vertical pairs: every one of the $$7$$ columns has $$6$$ adjacent pairs, again $$7 \times 6 = 42.$$

Total adjacent pairs $$= 42 + 42 = 84.$$

Therefore the total number of friends counted over all points is
$$\sum_{k=1}^{49} X(A_k) = 2 \times 84 = 168.$$

The expectation is the average per point:
$$E(X) = \frac{168}{49} = \frac{24}{7}.$$

Finally,
$$7E(X) = 7 \times \frac{24}{7} = 24.$$

Thus the required value is $$24$$.