1. Use the resistance formula

For a wire of length $L$, cross-sectional area $A$, and resistivity $\rho$,

$R = \rho \frac{L}{A}$

Since both wires have the same length and same thickness, they have the same $L$ and $A$.

Let

$k = \frac{L}{A}$

Then the resistances of the two wires are:

$R_1 = 6k, \qquad R_2 = 3k$

2. Find equivalent resistance in parallel

For two resistances in parallel,

$R_{\text{eq}} = \frac{R_1R_2}{R_1+R_2}$

Substitute $R_1=6k$ and $R_2=3k$:

$R_{\text{eq}} = \frac{(6k)(3k)}{6k+3k} = \frac{18k^2}{9k} = 2k$

3. Relate equivalent resistance to effective resistivity

If the combination is represented by an effective resistivity $\rho$, then

$R_{\text{eq}} = \rho \frac{L}{A} = \rho k$

But we already found

$R_{\text{eq}} = 2k$

So,

$\rho k = 2k$

$\rho = 2\ \Omega\text{ cm}$

4. Final answer

To the nearest integer,

$\boxed{2}$

5. Comparison with stored correct answer

Stored correct answer: $4$

Derived answer: $2$

These do not match.

The likely reason is that for equal length and area wires in parallel, the effective resistivity is not the arithmetic mean. It is obtained through parallel combination of resistances, giving

$\rho_{\text{eff}} = \frac{\rho_1\rho_2}{\rho_1+\rho_2} = \frac{6\times 3}{6+3} = 2$

So the stored answer appears incorrect.