Given:
- Original resistance \( R_1 = 5 \, \Omega \)
- The new length \( L_2 = 2L_1 \) (where \( L_1 \) is the original length)
- The new cross-sectional area \( A_2 = \frac{1}{2} A_1 \) (where \( A_1 \) is the original area)

 Step 1: Recall the formula for resistance

The resistance \( R \) of a wire is given by:

\(R = \rho \frac{L}{A}\)

where:
- \( \rho \) is the resistivity,
- \( L \) is the length,
- \( A \) is the cross-sectional area.

Step 2: Calculate the new resistance

The new resistance \( R_2 \) can be calculated using the new length and new area:

\(R_2 = \rho \frac{L_2}{A_2}\)

Substituting for \( L_2 \) and \( A_2 \):

\(R_2 = \rho \frac{2L_1}{\frac{1}{2}A_1} = \rho \frac{2L_1 \cdot 2}{A_1} = \rho \frac{4L_1}{A_1}\)

Step 3: Relate \( R_1 \) and \( R_2 \)

From the original resistance \( R_1 \):

\(R_1 = \rho \frac{L_1}{A_1}\)

Now, we can express \( R_2 \) in terms of \( R_1 \):

\(R_2 = 4 \left(\rho \frac{L_1}{A_1}\right) = 4R_1\)

Step 4: Substitute the value of \( R_1 \)

Now substitute \( R_1 = 5 \, \Omega \):

\(R_2 = 4 \times 5 \, \Omega = 20 \, \Omega\)

Therefore, the  new resistance is: \(\boxed{20 \, \Omega}\)