Step 1: Set up proportionalities.
Inside the wire the field is the potential gradient, \(E = V/l\), so \(E \propto V\) and \(E \propto 1/l\), while \(E\) is independent of the diameter. The resistance is \(R = \rho l / A\) with \(A = \pi d^2/4\), giving \(R = 4\rho l/(\pi d^2)\), so \(R \propto l\), \(R \propto 1/d^2\), and \(R\) is independent of \(V\). I will track only the ratios.
Step 2: Halving V.
Since \(E \propto V\), halving \(V\) halves the field: \(E \to E/2\). Since \(R\) does not contain \(V\) at all, the resistance stays the same. So field halves, resistance constant.
Step 3: Doubling l.
Since \(E \propto 1/l\), doubling \(l\) makes \(E \to E/2\). Since \(R \propto l\), doubling \(l\) makes \(R \to 2R\). So field halves and resistance doubles.
Step 4: Doubling d.
\(E\) has no \(d\) dependence, so the field is unaffected. Since \(R \propto 1/d^2\), replacing \(d\) by \(2d\) multiplies \(R\) by \(1/(2)^2 = 1/4\), so \(R \to R/4\). Field unchanged, resistance quartered.
Step 5: Collected results.
\[\boxed{(i)\ E \to \tfrac{E}{2},\ R\ \text{same};\quad (ii)\ E \to \tfrac{E}{2},\ R \to 2R;\quad (iii)\ E\ \text{same},\ R \to \tfrac{R}{4}}\]