How to open View from View Model?

You should not open a another view from ViewModel because it is against the MVVM pattern as "Hari p balineni" already said. So, instead of Creating Login Window, You can convert it into the UserControl and place it into your main window like

<v:logincontrol x:name="loginControl" height="270" width="560" xmlns:x="#unknown" xmlns:v="#unknown" />

and at code behind of the Login UserControl

public partial class LoginControl : UserControl
   {
       private UIElement parent;

       private string Username;
       private string Password;


       public ThreatBaseInformationTemplateV2()
       {
           InitializeComponent();
           this.Visibility = System.Windows.Visibility.Hidden;
       }

       public void SetParent(UIElement parent)
       {
           this.parent = parent;
       }

       public void ShowHandlerDailog()
       {
           this.Visibility = System.Windows.Visibility.Visible;
           this.parent.IsEnabled = false;
       }

       public void HideHandlerDialog()
       {
           this.Visibility = System.Windows.Visibility.Hidden;
           this.parent.IsEnabled = true;
       }

       private void btnLogin_Click(object sender, RoutedEventArgs e)
       {
           Username = txtUsername.Text;
           Password = txtPassword.Text;
           this.parent.IsEnabled = true;
           this.Visibility = System.Windows.Visibility.Hidden;
       }

       private void btnCancel_Click(object sender, RoutedEventArgs e)
       {
           this.parent.IsEnabled = true;
           this.Visibility = System.Windows.Visibility.Hidden;
       }
   }

Now, In the constructor of the MainWindow you can set the parent of this control like

this.loginControl.SetParent(KBEditor);

and the you can show this on LoadEvent of MainWindow.

Hope it will help.